Question: In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB = 6$ and $BC = 10$.  Find $\cos C$.
Solution: The triangle is shown below:

[asy]

pair A,B,C;

A = (0,0);

B = (6,0);

C = (0,8);

draw(A--B--C--A);

draw(rightanglemark(B,A,C,10));

label("$A$",A,SW);

label("$B$",B,SE);

label("$C$",C,N);

label("$10$",(B+C)/2,NE);

label("$6$",B/2,S);

[/asy]

The Pythagorean Theorem gives us $AC = \sqrt{BC^2 - AB^2} = \sqrt{100 - 36} = \sqrt{64}=8$, so $\cos C = \frac{AC}{BC} = \frac{8}{10} = \boxed{\frac45}$.